A particle executing simple harmonic motion with an amplitude A and angularr frequency `omega`. The ratio of maximum acceleration to the maximum velocity of the particle is

Options

A. `omegaA`
B. `omega^(2)A`
C. `omega`
D. `(omega^(2))/(A)`

Explanation

To find the ratio of maximum acceleration to the maximum velocity of a particle executing simple harmonic motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**:
A particle in simple harmonic motion can be described by its position as a function of time:
\[ x(t) = A \cos(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Calculate Velocity**:
The velocity \( v(t) \) of the particle is the derivative of the position with respect to time:
\[ v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t) \]
The maximum velocity \( v_{\text{max}} \) occurs when \( \sin(\omega t) = 1 \):
\[ v_{\text{max}} = A \omega \]

3. **Calculate Acceleration**:
The acceleration \( a(t) \) is the derivative of the velocity with respect to time:
\[ a(t) = \frac{dv}{dt} = -A \omega^2 \cos(\omega t) \]
The maximum acceleration \( a_{\text{max}} \) occurs when \( \cos(\omega t) = 1 \):
\[ a_{\text{max}} = A \omega^2 \]

4. **Find the Ratio**:
Now, we can find the ratio of maximum acceleration to maximum velocity:
\[ \text{Ratio} = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{A \omega^2}{A \omega} \]
Simplifying this gives:
\[ \text{Ratio} = \frac{\omega^2}{\omega} = \omega \]

5. **Conclusion**:
Therefore, the ratio of maximum acceleration to maximum velocity of the particle is:
\[ \text{Ratio} = \omega \]

### Final Answer:
The ratio of maximum acceleration to maximum velocity is \( \omega \).